Poisson sufficient statistics problem

Question

I have the following problem:

Let $Y_1, \dots, Y_n$ be a random sample from a Poisson distribution $\text{Pois}(\lambda)$. Recall, the $\text{Pois}(\lambda)$ distribution has the probability function $f_{\lambda}(y) = e^{-\lambda} \dfrac{\lambda^y}{y!}$, if $y = 0, 1, 2, 3, \dots$, and $\lambda > 0$.

(a) Show that $T(\mathbf{Y}) = \sum_{i = 1}^n Y_i$ is a sufficient statistic for $\lambda$ using the Fisher-Neyman factorisation theorem.

(b) What is the distribution of $T(\mathbf{Y})$? Obtain this result directly using the definition of a sufficient statistic.

For (a), we have that $L(\lambda, \mathbf{y}) = \prod_{i = 1}^n e^{-\lambda}\dfrac{\lambda^{y_i}}{y_i!} = e^{-n \lambda} \dfrac{\lambda^{\sum_{i = 1}^n y_i}}{\prod_{i = 1}^n y_i!}$. So $T(\mathbf{y}) = \sum_{i = 1}^n y_i$, $g(t, \lambda) = e^{-n \lambda} \lambda^t$ and $h(\mathbf{y}) = \dfrac{1}{\prod_{i = 1}^n y_i!}$.

For (b), the solution is given as follows:

$$T(\mathbf{Y}) \sim \text{Pois}(n \lambda)$$

$$P(\mathbf{Y} \mid T(\mathbf{Y})) = \dfrac{P(\mathbf{Y}, T(\mathbf{Y}))}{P(T(\mathbf{Y}))} = \dfrac{\prod_{i = 1}^n e^{-\lambda} \dfrac{\lambda^{Y_i}}{Y_i!}}{e^{-n \lambda}\dfrac{n^T \lambda^T}{T!}} = \dfrac{1}{n^T} \dfrac{T!}{\prod_{i = 1}^n Y_i!}$$

I don't understand the solution for (b). In particular, I don't understand how the author concluded that $T(\mathbf{Y}) \sim \text{Pois}(n \lambda)$ and $P(T(\mathbf{Y})) = e^{-n \lambda}\dfrac{n^T \lambda^T}{T!}$. I would greatly appreciate it if someone would please take the time to clarify this.

Answer 1

If $Y_1, ..., Y_n$ are iid $\text{Pois}(\lambda)$, then $T(Y) = \sum\limits_{i=1}^n Y_i \sim \text{Pois}(n\lambda)$

This is probably most easily seen by noting that the MGF for each $Y_i$ is $M_{y_i}(t) = e^{\lambda(e^t - 1)}$, so then the MGF for $T(Y)$ is $(M_{y_i}(t))^n = (e^{\lambda(e^t - 1)})^n = e^{(n\lambda)(e^t - 1)}$, which is the MGF for a $\text{Pois}(n\lambda)$ random variable.